# How do you find the center and vertex for -x^2 + y^2 – 2x – 2y – 9 = 0?

Jun 23, 2015

Centre (-1,1), vertices (-1,4) and (-1, -2)

#### Explanation:

Write the equation by completing the squares as ${\left(y - 1\right)}^{2} - {\left(x + 1\right)}^{2} = {3}^{2}$.

Or ${\left(y - 1\right)}^{2} / {3}^{2} - {\left(x + 1\right)}^{2} / {3}^{2} = 1$

This is an vertical hyperbola with centre at (-1,1)

Vertices would therefore be (-1, 1+3) and (-1, 1-3) that is (-1,4) and (-1,-2)